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TMUA Probability questions

4 original TMUA probability questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Conditional probability: restrict the outcomesProbability without replacement: update after each pick

Question 1TMUA Paper 1Warm-up

A bag contains 4 red counters and 8 blue counters. Two counters are taken out at random, one after the other, without replacement. What is the probability that they are both the same colour?
  1. A1766\dfrac{17}{66}
  2. B1633\dfrac{16}{33}
  3. C1733\dfrac{17}{33}
  4. D3566\dfrac{35}{66}
  5. E59\dfrac{5}{9}
Show the answer and worked solution

Answer: C · Method: Probability without replacement: update after each pick

  1. P(both red) =412×311= \dfrac{4}{12} \times \dfrac{3}{11} and P(both blue) =812×711= \dfrac{8}{12} \times \dfrac{7}{11}.
  2. That gives 1733\dfrac{17}{33}.

Why the tempting wrong answers are wrong

  • B: That's the probability of the opposite event.
  • E: That's with replacement. After the first counter is taken, there's one fewer in the bag.

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Question 2TMUA Paper 1Warm-up

A bag contains 2 red counters and 6 blue counters. Two counters are taken out at random, one after the other, without replacement. What is the probability that they are different colours?
  1. A314\dfrac{3}{14}
  2. B38\dfrac{3}{8}
  3. C1128\dfrac{11}{28}
  4. D37\dfrac{3}{7}
  5. E1328\dfrac{13}{28}
Show the answer and worked solution

Answer: D · Method: Probability without replacement: update after each pick

  1. P(red then blue) =28×67= \dfrac{2}{8} \times \dfrac{6}{7}, and P(blue then red) is the same.
  2. That gives 37\dfrac{3}{7}.

Why the tempting wrong answers are wrong

  • A: Red then blue and blue then red are both possible: add both orders.
  • B: That's with replacement. After the first counter is taken, there's one fewer in the bag.

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Question 3TMUA Paper 1Warm-up

A bag contains 2 red counters and 6 blue counters. Two counters are taken out at random, one after the other, without replacement. What is the probability that they are both red?
  1. A156\dfrac{1}{56}
  2. B128\dfrac{1}{28}
  3. C116\dfrac{1}{16}
  4. D114\dfrac{1}{14}
  5. E2728\dfrac{27}{28}
Show the answer and worked solution

Answer: B · Method: Probability without replacement: update after each pick

  1. P(both red) =28×17= \dfrac{2}{8} \times \dfrac{1}{7}.
  2. That gives 128\dfrac{1}{28}.

Why the tempting wrong answers are wrong

  • C: That's with replacement. After the first counter is taken, there's one fewer in the bag.
  • E: That's the probability of the opposite event.

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Question 4TMUA Paper 2Standard

Two fair six-sided dice are rolled. Given that the total is 77, what is the probability that the product of the two scores is even?
  1. A13\dfrac{1}{3}
  2. B12\dfrac{1}{2}
  3. C23\dfrac{2}{3}
  4. D56\dfrac{5}{6}
  5. E11
Show the answer and worked solution

Answer: E · Method: Conditional probability: restrict the outcomes

  1. An odd total means one score is odd and the other is even.
  2. So the product always contains an even factor: the probability is 11.

Why the tempting wrong answers are wrong

  • B: Think about what an odd total forces: one die must be even.

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